- #1
rxh140630
- 60
- 11
- Homework Statement
- prove lg(n!) = thetha(nlgn) where lg n is log with base = 2. Use Sterling's approximation as a hint
- Relevant Equations
- Sterling's approximation: [itex]n! = \sqrt(2*pi*n)(\frac{n}{e})^n(1+ \theta{1/n}) [/itex]
Sterling's approximation: [itex]n! = \sqrt{2*pi*n}(\frac{n}{e})^n(1+ \theta(1/n)) [/itex]
So I need to prove
[itex] c_1nlgn ≤lg(\sqrt{2*pi*n}) + lg((\frac{n}{e})^n) + lg(1+ \theta(1/n))) ≤ c_2nlgn[/itex]
My question is:
assume I've proven [itex]lg(\sqrt{2*pi*n})[/itex] as [itex] \theta(lgn) [/itex]
Do I need to now prove that [itex] c_1nlgn ≤ lgn ≤ c_2nlgn [/itex] ??
What if we assume I found that [itex]lg(\sqrt{2*pi*n})[/itex] as [itex] \theta(\sqrt{n}) [/itex]
Do I need to now prove that [itex] c_1nlgn ≤ \sqrt{n} ≤ c_2nlgn [/itex] ??
Assuming the other terms in the sum [itex] g(\sqrt{2*pi*n}) + lg((\frac{n}{e})^n) + lg(1+ \theta(1/n))) [/itex] are proven to be [itex] \theta(nlgn) [/itex]
So I need to prove
[itex] c_1nlgn ≤lg(\sqrt{2*pi*n}) + lg((\frac{n}{e})^n) + lg(1+ \theta(1/n))) ≤ c_2nlgn[/itex]
My question is:
assume I've proven [itex]lg(\sqrt{2*pi*n})[/itex] as [itex] \theta(lgn) [/itex]
Do I need to now prove that [itex] c_1nlgn ≤ lgn ≤ c_2nlgn [/itex] ??
What if we assume I found that [itex]lg(\sqrt{2*pi*n})[/itex] as [itex] \theta(\sqrt{n}) [/itex]
Do I need to now prove that [itex] c_1nlgn ≤ \sqrt{n} ≤ c_2nlgn [/itex] ??
Assuming the other terms in the sum [itex] g(\sqrt{2*pi*n}) + lg((\frac{n}{e})^n) + lg(1+ \theta(1/n))) [/itex] are proven to be [itex] \theta(nlgn) [/itex]